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(B) The addition of $HI$ to alkenes in the presence of peroxide does not follow the anti-Markovnikov rule because the free radical mechanism is not fe

Vedclass · Accepted Answer

$2-$Iodopropane

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(B) The addition of $HI$ to alkenes in the presence of peroxide does not follow the anti-Markovnikov rule because the free radical mechanism is not feasible for $HI$ due to the endothermic nature of the iodine radical formation step.Therefore,the reaction follows the Markovnikov addition rule,where the electrophile $H^+$ adds to the carbon with more hydrogen atoms.$CH_3-CH=CH_2 + HI \rightarrow CH_3-CH(I)-CH_3$The product formed is $2-$iodopropane (also known as isopropyl iodide).

The name of the product obtained by the addition of $HI$ to propene in the presence of a peroxide catalyst is:

Similar Questions

Which of the following reactions is correct?

Which of the following alkenes,on oxidation by $KMnO_4$ in dil. $H_2SO_4$,forms adipic acid?

The addition of $HI$ to the double bond of propene yields isopropyl iodide and not $n$-propyl iodide because the addition proceeds through the formation of:

$I$. $(CH_3)_2 C=CH_2 \xrightarrow{KMnO_4 / H^{+}} X + CO_2 + H_2 O$
$II$. $CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4 / H^{+}} Y$
The functional groups in $X$ and $Y$ are respectively

The major product formed in the following reaction is:

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